# Math Help - Perms and combs

1. ## Perms and combs

In how many ways can 8 oarsmen be seated in an 8 oared boat when only 3 can be on the stroke side and only 3 can row on the bow side.

answer is 1152, but how do we i get this, thank you.

2. Maybe I'm missing something but where do the other 2 oarspeople () sit? Does order matter? I'll assume not.

First ask, how many ways are there of picking the first 3? Answer:
${8\choose3}$

Now for each one of those you have 5 different people left. How many ways are there to pick the next 3? Answer:
${5\choose3}$

Now for each of those you have 2 people left and 2 spots (whether they oar or not), but there is only 1 way to pick them.
${2\choose2}=1$

So:
${8\choose3}{5\choose3}=560$

So either I am understanding the problem wrong or the answer you have is wrong. I don't think I'm doing it wrong based on my interpretation but you never know...

3. ## Permutations

hmmm im not sure if the answer is wrong or your wrong because i did pretty much the same method as you did. This question i think its mainly talking bout permutation rather than combination........ im not sure

4. Well if its permutation (i.e. the three people on the left siting in a different order count differently, same with the right) then we get:

$\mathbb{P}^8_3\mathbb{P}^5_3 = 20160$

And I still don't know what the other two oarsmen are doing.

The above answer is for the case where they are just sitting off to the side (so order doesn't matter for them). If order matters for all of them, then the answer is just $8!=40320$

You know actually, the whole problem isn't stated very well, the bow is the front and the stroke side is the port or leftside of the boat. So oarsmen are either in the front or the left?

And really, if it is permutation, then the whole business about left or right or front or back is a total red herring. Every spot is different so it wouldn't matter where the spots were.

Just for giggles I tried factoring their answer to see if I could make any sense of it. $1152 = 2^{7} \cdot 3^{2} = 1 \cdot 2 \cdot 3 \cdot 4 \cdot 6 \cdot 8$

I can't see how that works into the problem at all. My guess is that, unless you didn't put some crucial piece of information out of the original post, the answer is wrong or 1152 is the answer for another problem?. Believe it or not, it happens a lot.

I say again: